# Talk:Centrifugal force

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## Centrifugal force ​mv2⁄r

Neither vc=​mv2r nor ac=​v2r is anywhere in this article. I feel that at least one of them should be somewhere. — Preceding unsigned comment added by 72.172.59.119 (talk) 22:05, 24 July 2016‎

There not in this article because Fc and ac are the centripetal force and acceleration. --FyzixFighter (talk) 13:05, 29 July 2016 (UTC)

## Velocity proof

In the velocity derivation, the wording and formulæ are contradictory in terms of what applies to rotating versus stationary reference frames. —DIV (120.19.177.139 (talk) 02:45, 6 September 2016 (UTC))

## direction of centrifugal force

Regarding this edit, asserting that:

ω x (ω x r) is always perpendicular to ω

I don't think this is true.

If F_centrifugal = mω x (ω x r)

then by the rules of the [triple product]:

=F_centrifugal=m(${\displaystyle \mathbf {\omega } (\mathbf {\omega } \cdot \mathbf {r} )-\mathbf {r} (\mathbf {\omega } \cdot \mathbf {\omega } )}$ =m(${\displaystyle \mathbf {\omega } (\mathbf {\omega } \cdot \mathbf {r} )-\mathbf {r} \omega ^{2}}$).

Yes, the second term is always radially outward, since it has magnitude ${\displaystyle m*\omega ^{2}}$ in the r direction, but the first time doesn't have to be 0 so long as ${\displaystyle \omega }$ and r are not perpendicular.

For example, consider an objective in helical motion such that it is completing counterclockwise circles in xy plane as viewed from +z direction so that its r has a component n the z direction so that ${\displaystyle \omega }$ has a component in the xy plane. That would make ${\displaystyle \omega \cdot \mathbf {r} \neq 0}$ and thus ω x (ω x r) has a component in the ${\displaystyle \mathbf {\omega } }$ direction and is not only radially outward.--Louiedog (talk) 11:43, 5 June 2017 (UTC)

I have just realised that the formula for centrifugal force you quote from the article was missing a minus sign. But with that correction, yes indeed, Fcentrifugal = – m {(ω · r) ω – (ω · ω) r} , and therefore
ω · Fcentrifugal = – m {(ω · r) (ω · ω) – (ω · ω) (ω · r)} = 0 .
That is, Fcentrifugal is perpendicular to ω. It is, in fact, a well-known property of the vector cross product, a × b, that it is always perpendicular to both of its multiplicands, a and b, and, in particular, –ω × (ω × r) is always perpendicular to ω. It is, in fact, just ω2 times the projection of the vector r onto the plane perpendicular to ω.
Note also that the centrifugal force is an artefact of the motion of the coordinate system in which the motion of objects is being described, and this does not necessarily bear any relation whatsoever to the motion of any of those objects themselves. I presume the helical motion you have in mind is one in which the body's position r(t) at time t is given by something like r(t) = ρ cos(ω t) i + ρ sin(ω t) j + vz t k. But the ω in this expression is the angular velocity of the body in question about the z axis, and not (necessarily) that of the coordinate system in which the motion of the body is being described. If the coordinate system determined by the three unit vectors i, j and k is inertial, for instance, then the angular velocity of that coordinate sytem is zero, and there will be no centrifugal force on the body in that coordinate system. For the body to be following such a helical path, there must, of course, be a proper force,
m (t) = − m ω2 ( ρ cos(ω t) i + ρ sin(ω t) j ) ,
acting on it. This force is directed towards, and perpendicular to, the z axis of the inertial coordinate system. In a coordinate system whose origin and z axis coincide with those of the inertial system, but is rotating with angular velocity ω = ω k, the x and y coordinates of the body will be constant, and it will be moving with uniform velocity parallel to the z axis. In this coordinate system there will be a centrifugal force m ω2 ( ρ cos(ω t) i + ρ sin(ω t) j ) acting on the body which exactly balances the proper force. This centrifugal force is clearly, as it must always be, perpendicular to ω.
David Wilson (talk · cont) 14:21, 5 June 2017 (UTC)